Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A block A of mass $m_1$ rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of the table and from its other end, another block B of mass $m_2$ is suspended. The coefficient of kinetic friction between block A and the table is $\mu_k$ . When block A is sliding on the table, the tension in the string is:

$\frac{(m_2 + \mu_k m_1)g}{(m_1 + m_2)}$

$\frac{(m_2 - \mu_k m_1)g}{(m_1 + m_2)}$

$\frac{m_1 m_2 (1 - \mu_k)g}{(m_1 + m_2)}$

$\frac{m_1 m_2 (1 + \mu_k)g}{(m_1 + m_2)}$

Step-by-Step Solution

System Analysis: Block B ( $m_2$ ) moves vertically downwards and block A ( $m_1$ ) moves horizontally. Both share the same magnitude of acceleration $a$ because they are connected by an inextensible string.
Equations of Motion:

For Block B ( $m_2$ ): The forces are weight $m_2g$ (down) and tension $T$ (up). Since it accelerates down: $m_2g - T = m_2a \quad \text{...(i)}$
For Block A ( $m_1$ ): The forces are tension $T$ (forward) and kinetic friction $f_k$ (backward). The vertical forces balance ( $N = m_1g$ ), so $f_k = \mu_k N = \mu_k m_1g$ . Since it accelerates forward: $T - f_k = m_1a \Rightarrow T - \mu_k m_1g = m_1a \quad \text{...(ii)}$

Finding Acceleration: Add equations (i) and (ii) to eliminate $T$ : $m_2g - \mu_k m_1g = (m_1 + m_2)a$ $a = \frac{g(m_2 - \mu_k m_1)}{m_1 + m_2}$
Finding Tension: Substitute $a$ back into equation (ii): $T = m_1a + \mu_k m_1g$ $T = m_1 \left[ \frac{g(m_2 - \mu_k m_1)}{m_1 + m_2} \right] + \mu_k m_1g$ $T = \frac{m_1 m_2 g - \mu_k m_1^2 g + \mu_k m_1^2 g + \mu_k m_1 m_2 g}{m_1 + m_2}$ $T = \frac{m_1 m_2 g (1 + \mu_k)}{m_1 + m_2}$ (Reference: NCERT Class 11, Physics Part I, Chapter 5, Section 5.10 and Example 5.9).

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